What is the integral from 1 to 3 of (x+3)/(x^2+6x) ?

The integral from 1 to 3 of (x+3)/(x^2+6x) is (3ln(3)-ln(7))/2

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integral from 1 to 3 of (x+3)/(x^2+6x)

\int_{1}^{3} \frac{x + 3}{x ^{2} + 6 x} d x

\frac{3 ln ( 3 ) - ln ( 7 )}{2}

Decimal Notation

0.67496 \dots

Solution steps

\int_{1}^{3} \frac{x + 3}{x ^{2} + 6 x} d x

Take the partial fraction of

\frac{x + 3}{x ^{2} + 6 x} : \frac{1}{2 x} + \frac{1}{2 ( x + 6 )}

= \int_{1}^{3} \frac{1}{2 x} + \frac{1}{2 ( x + 6 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{3} \frac{1}{2 x} d x + \int_{1}^{3} \frac{1}{2 ( x + 6 )} d x

\int_{1}^{3} \frac{1}{2 x} d x = \frac{1}{2} ln (3)

\int_{1}^{3} \frac{1}{2 ( x + 6 )} d x = ln (3) - \frac{1}{2} ln (7)

= \frac{1}{2} ln (3) + ln (3) - \frac{1}{2} ln (7)

Simplify

\frac{1}{2} ln (3) + ln (3) - \frac{1}{2} ln (7) : \frac{3 ln ( 3 ) - ln ( 7 )}{2}

= \frac{3 ln ( 3 ) - ln ( 7 )}{2}

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What is the integral from 1 to 3 of (x+3)/(x^2+6x) ?
The integral from 1 to 3 of (x+3)/(x^2+6x) is (3ln(3)-ln(7))/2